Probability Questions

Solution : Required Probability = P(X = 0) + P(X = 1) = \(e^{-5}\over 0!\).\(5^0\) + \(e^{-5}\over 1!\).\(5^1\) = \(e^{-5}\) + 5\(e^{-5}\) = \(6\over {e^5}\) Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence from match to match. The probability that in a match series India’s […]

Solution : Let the events, A = Ist aeroplane hit the target B = 2nd aeroplane hit the target And their corresponding probabilities are P(A) = 0.3 and P(B) = 0.2 \(\implies\) P(A’) = 0.7 and P(B’) = 0.8 \(\therefore\)  Required Probability = P(A’)P(B) + P(A’)P(B’)P(A’)P(B) + ……. = (0.7)(0.2) + (0.7)(0.8)(0.7)(0.2) + ……. =

Solution : Probability of getting score 9 in a single throw = \(4\over 36\) = \(1\over 9\) Probability of getting score 9 exactly in double throw = \(^3C_2\) \(\times\) \(({1\over 9})^2\) \(\times\) \(8\over 9\) = \(8\over 243\) Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence

Solution : A = {4. 5. 6}  and  B = {1, 2, 3, 4} \(A \cap B\) = 4 \(\therefore\)  \(P(A \cup B)\) = P(A) + P(B) – \(A \cap B\) \(\implies\) \(P(A \cup B)\) = 3/6 + 4/6 – 1/6 = 1 Similar Questions The probability of India winning a test match against the

Solution : We know that, P(A/B) = \(P(A \cap B)\over P(B)\)    …….(i) and P(B/A) = \(P(B \cap A)\over P(A)\)    ……….(ii) \(\therefore\)  P(B) = \(P(B/A).P(A)\over P(A/B)\) = \(1\over 3\) Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence from match to match. The probability that

Solution : S = { 00, 01, 02, ……, 49 } Let A be the event that sum of the digits on the selected ticket is 8, then A = { 08, 17, 26, 35, 44 } Let B be the event that the product of the digits is zero. B = { 00, 01,

Solution : Total number of cases = \(^9C_3\) = 84 Number of favourable cases = \(^3C_1\).\(^4C_1\).\(^2C_1\) = 24 \(\therefore\)  P = \(24\over 84\) = \(2\over 7\) Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence from match to match. The probability that in a match series

Solution : \(P({A’ \cap B’\over C})\) = \(P(A’ \cap B’ \cap C)\over P(C)\) = \(P(C) – P(A \cap C) – P(B \cap C) + P(A \cap B\cap C)\over P(C)\)   ……..(i) Given, \(P(A \cap B\cap C)\)  = 0 and A, B and C are pairwise independent. \(\therefore\)  \(P(A \cap C)\) = P(A).P(C) and \(P(B \cap C)\)

Solution : Probability of guessing a correct answer, p = \(1\over 3\) and probability of guessing a wrong answer, q  = \(2\over 3\) So, the probability of guessing 4 or more correct answers is = \(^5C_4\) \(({1\over 3})^4\). \(2\over 3\) + \(^5C_5\) \(({1\over 3})^5\) = \(5.2\over {3^5}\) + \(1\over {3^5}\) = \(11\over {3^5}\) Similar Questions

Question : If C and D are two events such that C \(\subset\) D and P(D) \(\ne\) 0, then the correct statement among the following is (a) P(C/D) \(\ge\) P(C) (b) P(C/D) < P(C) (c) P(C/D) = \(P(D)\over P(C)\) (d) P(C/D) = P(C) Solution : As P(C/D) = \(P(C \cap D)\over P(D)\) = \(P(C)\over P(D)\)